Praeclarum Theorema/Formulation 1/Proof 1
Theorem
- $\paren {p \implies q} \land \paren {r \implies s} \vdash \paren {p \land r} \implies \paren {q \land s}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\paren {p \implies q} \land \paren {r \implies s}$ | Premise | (None) | ||
| 2 | 1 | $p \implies q$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
| 3 | 1 | $r \implies s$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
| 4 | 4 | $p \land r$ | Assumption | (None) | ||
| 5 | 4 | $p$ | Rule of Simplification: $\land \EE_1$ | 4 | ||
| 6 | 4 | $r$ | Rule of Simplification: $\land \EE_2$ | 4 | ||
| 7 | 1, 4 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 5 | ||
| 8 | 1, 4 | $s$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 3, 6 | ||
| 9 | 1, 4 | $q \land s$ | Rule of Conjunction: $\land \II$ | 7, 8 | ||
| 10 | 1 | $\paren {p \land r} \implies \paren {q \land s}$ | Rule of Implication: $\implies \II$ | 4 – 9 | Assumption 4 has been discharged |
$\blacksquare$